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Problem 990

AMC 12 late, AIME early

Algebra Difficulty 4.9 Prove it Berkeley Math Circle Monthly Contest 3 · United States

Let $0 < a_{0} \leq a_{1} \leq \cdots \leq a_{n}$ . If $z$ is a complex number such that $a_{0} z^{n} + a_{1} z^{n-1} + \cdots + a_{n} = 0$ prove that $|z| \geq 1$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Solution:

Assume that $|z| < 1$ . If $a_{0} z^{n} + a_{1} z^{n-1} + \cdots + a_{n} = 0$ then $a_{0} z^{n+1} + a_{1} z^{n} + \cdots + a_{n} z = 0$ and subtracting these two equations leads to $a_{0} z^{n+1} + (a_{1} - a_{0}) z^{n} + \cdots + (a_{n} - a_{n-1}) z - a_{n} = 0$ , or equivalently $a_{n} = a_{0} z^{n+1} + (a_{1} - a_{0}) z^{n} + \cdots + (a_{n} - a_{n-1}) z$ hence

$\begin{aligned} |a_{n}| & = \left| a_{0} z^{n+1} + (a_{1} - a_{0}) z^{n} + \cdots + (a_{n} - a_{n-1}) z \right| \\ & \leq a_{0} |z|^{n+1} + (a_{1} - a_{0}) |z|^{n} + \cdots + (a_{n} - a_{n-1}) |z| \\ & < a_{0} + (a_{1} - a_{0}) + \cdots + (a_{n} - a_{n-1}) = a_{n}, \end{aligned}$

which is impossible. Thus $|z| \geq 1$ .

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty, ordering) added by this project.